While riding in a hot air balloon, which is steadily descending at a speed of 1.39 m/s, you accidentally drop your cell phone.

(a) After 4.00 s, what is the speed of the cell phone?

(b) How far is the cell phone below the balloon after this time?
d = m

(c) What are your answers to parts (a) and (b) if the balloon is rising steadily at 1.39 m/s?
v =

The analysis is the same as in parts (a) and (b), only now the balloon is moving upward. How will that affect the distance between the balloon and phone at the later time? m/s
d = m

To answer these questions, we need to consider the motion of the cell phone relative to the hot air balloon.

(a) After 4.00 s, what is the speed of the cell phone?

The speed of the cell phone will remain constant since it is dropped from the balloon and not subjected to any horizontal forces. Therefore, its speed will be the same as the balloon's descent speed, which is 1.39 m/s.

(b) How far is the cell phone below the balloon after this time?

To calculate the distance the cell phone has fallen, we can use the formula:

distance = initial velocity * time + (1/2) * acceleration * time^2

Since the phone is dropped initially, its initial velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2 (assuming no air resistance). Plugging in the values:

distance = 0 * 4 + (1/2) * 9.8 * (4)^2 = 78.4 m

Therefore, the cell phone is 78.4 meters below the balloon after 4.00 seconds.

(c) What are your answers to parts (a) and (b) if the balloon is rising steadily at 1.39 m/s?

If the balloon is rising steadily at 1.39 m/s, the analysis for parts (a) and (b) remains the same. The speed of the cell phone will still be 1.39 m/s, and the distance it falls will still be 78.4 meters.

However, in this case, the distance between the balloon and the phone will increase over time, since the balloon is moving upward. The phone will still fall due to gravity, but the upward motion of the balloon will reduce the rate at which the phone moves away from the balloon.